You forgot that the capacitor's reactance is 90 degrees out of phase with the resistance. You have to find the vector magnitude of the denominator.
Gain = 20 log( (1/wc) / sqrt((R^2) + (1/wc)^2)Gain = -20.05dB

I got 198.9 mV as Vo, which is approximately -20.05 dB with respect to 2 Vi.

I didn't understand your Vo being -14.03 dB

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The OP wrote:

(Vout/Vin) = (1/jwc) / ((1/jwc)+R)

R = 1khz
C=3.1847 x 10^-8
f= 50khz
Vin= 2v

Find Vout express in Db

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You got Vout=198.9 mV.
20 log (.1989) = -14.03dB.
You are correct in showing gain as -20.05dB, which will of course be true for any input voltage. I realize that dB is a unitless gain measurement (a ratio), but the OP said

Find Vout express in Db.

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I suppose to be correct, we should say
Vout=-14.03dBV, i.e., relative to 1 volt, which identifies it as an absolute measurement rather than a ratio.